(x^2-49)^2+(x^2+4x-21)^2=0

3 min read Jun 17, 2024

Solving the Equation: (x^2-49)^2 + (x^2+4x-21)^2 = 0

This equation presents a unique challenge due to the presence of squared terms. Let's break down the steps involved in finding its solution.

Understanding the Problem

The equation (x^2-49)^2 + (x^2+4x-21)^2 = 0 implies that the sum of two squares is equal to zero. A fundamental principle in mathematics states that the square of any real number is always greater than or equal to zero. Therefore, the only way for the sum of two squares to equal zero is if both squares are individually equal to zero.

Solving for x

Set each squared term to zero:
- (x^2-49)^2 = 0
- (x^2+4x-21)^2 = 0
Take the square root of both sides:
- x^2 - 49 = 0
- x^2 + 4x - 21 = 0
Solve the resulting quadratic equations:
- For x^2 - 49 = 0:
  - x^2 = 49
  - x = ±7
- For x^2 + 4x - 21 = 0:
  - (x + 7)(x - 3) = 0
  - x = -7 or x = 3

Solution and Verification

We have found two possible solutions for x: x = 7 and x = -7. However, when we substitute these values back into the original equation, we find that only x = -7 satisfies the equation.

Verification:

For x = -7:
- (-7^2 - 49)^2 + (-7^2 + 4(-7) - 21)^2 = 0
- (49 - 49)^2 + (49 - 28 - 21)^2 = 0
- 0 + 0 = 0 (This confirms the solution)
For x = 7:
- (7^2 - 49)^2 + (7^2 + 4(7) - 21)^2 = 0
- (49 - 49)^2 + (49 + 28 - 21)^2 = 0
- 0 + 56^2 ≠ 0 (This does not satisfy the equation)

Conclusion

Therefore, the only solution to the equation (x^2-49)^2 + (x^2+4x-21)^2 = 0 is x = -7.